Question

Let $\alpha$ and $\beta$ be the roots of equation $x^2-6x-2=0.$ If $a_n={\alpha }^n-{\beta }^n,$ for $n\ge 1$ , then the value of $\frac{a_{10}-2a_8}{2a_9}$ is

• A. 6
• B. -6
• C. 3
• D. -3

Answer 1

Answer: (C)

$x^2-6x-2=0{<}_{\beta }^{\alpha }$
$\Rightarrow {\alpha }^2-6\alpha -2=0$
${\beta }^2-6\beta -2=0\dots ..(1)$
$a_n={\alpha }_n-{\beta }_{n}n\ge 1$
$a_{40}-29_8={\alpha }^{10}-{\beta }^{10}-2{\alpha }^8+2{\beta }^8$
$={\alpha }^8\left(a^2-2\right)-b^8\left(b^2-2\right)$
$={\alpha }^8(6\alpha )-{\beta }^8(6\beta )$ (using (1)
$=6{\alpha }^9-6{\beta }^9$
$=6a_9$
now
$\frac{a_{10}-2a_8}{2a_9}=\frac{6a_9}{2a_9}=3$